Exercises — Acceleration of rolling objects on inclines — comparison

The shape factor is defined as $\beta =I/(m{R}^2)$ — you literally read off the number in front of $m{R}^2$. Here $I=\frac{2}{5}m{R}^2$, so $\beta =\frac{\frac{2}{5}m{R}^2}{m{R}^2}=\frac{2}{5}=\mathrm{0.40.}$ A $\beta$ of $2/5$ is the solid sphere. ✔

Acceleration is $a=\frac{g\sin \theta }{1+\beta }$. A larger $\beta$ sits in the denominator, so it makes $a$ smaller. The sphere has the smaller $\beta$ ($2/5<1$), so the larger $a$, so it wins. Solid sphere first. ✔

Solid cylinder $\Rightarrow \beta =\frac{1}{2}$, so $1+\beta =\frac{3}{2}$. $a=\frac{g\sin \theta }{1+\beta }=\frac{9.8\times \sin 25^{\circ }}{1.5}.$ Now $\sin 25^{\circ }=0.4226$, so the numerator is $9.8\times 0.4226=4.141$. $a=\frac{4.141}{1.5}=2.76{\text{m/s}}^2.$ Check the size: it must be less than the free-slide value $g\sin \theta =4.14$ — and $2.76<4.14$. ✔

Use the energy result built above in engine (b): dropping $h$ releases $mgh$, which splits into forward motion $\frac{1}{2}m{v}^2$ and spin $\frac{1}{2}I{\omega }^2=\frac{1}{2}\beta m{v}^2$, giving $mgh=\frac{1}{2}m{v}^2(1+\beta )$ and hence $v=\sqrt{\frac{2gh}{1+\beta }}.$ Solid sphere $\Rightarrow \beta =\frac{2}{5}$, so $1+\beta =\frac{7}{5}$. $v=\sqrt{\frac{2(9.8)(2.0)}{1.4}}=\sqrt{\frac{39.2}{1.4}}=\sqrt{28.0}=5.29\text{m/s}.$ Sanity: a frictionless sliding block would reach $\sqrt{2gh}=\sqrt{39.2}=6.26$ m/s — the roller is slower because some energy went into spin (the $(1+\beta )$ tax). $5.29<6.26$. ✔

We need friction in terms of $a$. Friction acts at the rim (distance $R$ from the centre), so its torque is $fR$; Newton's rotational law sets that equal to $I\alpha$ — the spin-inertia $I$ times the angular acceleration $\alpha$ (defined at the top: $\alpha =a/R$, the spin-rate change tied to the centre's acceleration). Substituting $I=\beta m{R}^2$ and $\alpha =a/R$, exactly as in engine (a): $fR=I\alpha =(\beta m{R}^2)\frac{a}{R}=\beta mRa\Rightarrow f=\beta ma.$ First get $a$ for the cylinder ($\beta =\frac{1}{2}$): $a=\frac{g\sin \theta }{1+\beta }=\frac{9.8\times 0.5}{1.5}=\frac{4.9}{1.5}=3.267{\text{m/s}}^2.$ Then $f=\beta ma=\frac{1}{2}(3.0)(3.267)=4.90\text{N}.$ Look at figure s01: the yellow $f$ points up the slope even though the body slides down — because friction's job is to supply the red torque arc ($fR$) that spins the cylinder up. ✔

Static friction can supply at most $\mu N$, where $N=mg\cos \theta$ is the normal force defined at the top of this page — the slope's perpendicular push, drawn as the blue arrow in figure s01. Rolling holds only while the required friction $f=\beta ma$ does not exceed this cap: $\beta ma\le \mu mg\cos \theta .$ The mass cancels. Rearranging with $a=\frac{g\sin \theta }{1+\beta }$:

Each has constant acceleration $a=\frac{g\sin \theta }{1+\beta }$, starting from rest, so distance obeys $s=\frac{1}{2}a{t}^2$. Compute the accelerations. Common factor $g\sin 20^{\circ }=9.8\times 0.3420=3.352$.

• Solid sphere $\beta =\frac{2}{5}$, $1+\beta =1.4$: $a_1=3.352/1.4=2.394{\text{m/s}}^2$.
• Hollow sphere $\beta =\frac{2}{3}$, $1+\beta =1.6667$: $a_2=3.352/1.6667=2.011{\text{m/s}}^2$.

Time for the solid sphere to cover $L=4.0$ m: $L=\frac{1}{2}{a}_1{t}^2\Rightarrow t=\sqrt{\frac{2L}{a_1}}=\sqrt{\frac{8.0}{2.394}}=\sqrt{3.342}=1.828\text{s}.$ At that same instant the hollow sphere has travelled $s_2=\frac{1}{2}{a}_2{t}^2=\frac{1}{2}(2.011)(1.828^2)=\frac{1}{2}(2.011)(3.342)=3.361\text{m}.$ So the gap behind is $\Delta =L-s_2=4.0-3.361=0.639\approx 0.64\text{m}.$ The hollow sphere trails by about 0.64 m when the solid one finishes. ✔

From $L=\frac{1}{2}a{t}^2$ with $L$ fixed: $t=\sqrt{2L/a}\propto \sqrt{1/a}\propto \sqrt{1+\beta }$ (since $a\propto 1/(1+\beta )$). Therefore

Invert the energy result $v=\sqrt{\frac{2gh}{1+\beta }}$ (engine (b)). Square both sides: $v^2=\frac{2gh}{1+\beta }\Rightarrow 1+\beta =\frac{2gh}{v^2}.$ Plug in: $2gh=2(9.8)(1.5)=29.4$ and $v^2=4.2^2=17.64$. $1+\beta =\frac{29.4}{17.64}=1.6667\Rightarrow \beta =0.6667=\frac{2}{3}.$ A $\beta$ of $2/3$ is the hollow sphere (thin shell). ✔

Pure rolling survives while the friction the surface can give ($\mu$) meets or beats the friction required: $\mu \ge \frac{\beta \tan \theta }{1+\beta }$. The steepest allowed angle is when they are equal: $\mu =\frac{\beta \tan {\theta }_{\max }}{1+\beta }\Rightarrow \tan {\theta }_{\max }=\frac{\mu (1+\beta )}{\beta }.$ For the ring $\beta =1$, $1+\beta =2$: $\tan {\theta }_{\max }=\frac{0.25\times 2}{1}=0.5\Rightarrow {\theta }_{\max }=\arctan (0.5)=26.57^{\circ }.$ Steeper than $\approx 26.6^{\circ }$ and the ring starts to skid. ✔

Take $\beta \to 0$ in $a=\frac{g\sin \theta }{1+\beta }$: ${\lim }_{\beta \to 0}a=\frac{g\sin \theta }{1+0}=g\sin \theta .$ That is exactly the sliding block on a frictionless incline. Physically $\beta =I/m{R}^2\to 0$ means no rotational inertia — a point mass, or all mass on the axis — so nothing has to be spun up and all of gravity goes to translation. Likewise $v=\sqrt{2gh/(1+\beta )}\to \sqrt{2gh}$, the free-fall/slide speed. The rolling law contains the sliding law as its zero-spin corner. ✔

• Friction on a roller, in terms of $a$? ::: $f=\beta ma$
• What is $\alpha$, and how does it link to $a$? ::: angular acceleration; $\alpha =a/R$
• What is $N$ on an incline? ::: normal force $=mg\cos \theta$
• Minimum $\mu$ for pure rolling? ::: ${\mu }_{\min }=\frac{\beta \tan \theta }{1+\beta }$
• Steepest rolling angle? ::: ${\theta }_{\max }=\arctan (\mu (1+\beta )/\beta )$
• Recover $\beta$ from a measured bottom speed? ::: $1+\beta =\frac{2gh}{v^2}$
• Descent-time ratio of two shapes? ::: $\frac{t_2}{t_1}=\sqrt{\frac{1+{\beta }_2}{1+{\beta }_1}}$